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Solution o differential equation x\frac{\mathrm{d} y}{\mathrm{d} x} = y (\ln y -\ln x + 1) is

  • Option 1)

    y = -x\cdot e^{cx}

  • Option 2)

    y = 2x\cdot e^{cx}

  • Option 3)

    y = x\cdot e^{cx}

  • Option 4)

    y = -2x\cdot e^{cx}

 

Answers (1)

best_answer

As we have learnt,

 

General form of Variable Separation -

d\left (log \frac{y}{x} \right )=\frac{xdy-ydx}{xy}

-

 

 Given equation can we written as -

x \frac{d y }{d x} = y \ln \left (\frac{y}{x} \right ) + y \\*\Rightarrow \frac{xdy - ydx}{dx} = y\ln \left(\frac{y}{x} \right ) \Rightarrow xdy -ydx = y\ln \left(\frac{y}{x} \right )\cdot dx

Divide through out by 'xy', we get

\frac{xdy-ydx}{xy} = \frac{\ln\left(\frac{y}{x} \right )}{x}dx \\*\Rightarrow \frac{d\left(\ln\left(\frac{y}{x} \right ) \right )}{\ln\left(\frac{y}{x} \right )} = \frac{dx}{x}

On integrating it gives,

\\*\Rightarrow\ln\left(\frac{\ln\left(\frac{y} {x}\right )}{x} \right ) = x \Rightarrow \frac{\ln\left(\frac{y}{x} \right )}{x} = e^c \\*\Rightarrow \frac{\ln\left(\frac{y}{x} \right )}{x} = C\Rightarrow \ln\left(\frac{y}{x} \right ) = cx \Rightarrow \frac{y}{x} = e^{Cx} \Rightarrow y = x\cdot e^{cx}

 


Option 1)

y = -x\cdot e^{cx}

Option 2)

y = 2x\cdot e^{cx}

Option 3)

y = x\cdot e^{cx}

Option 4)

y = -2x\cdot e^{cx}

Posted by

Himanshu

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